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When I was reading 'Convex Optimization, Stephen Boyd', I was wondering of following steps

  • Consider the unit circle in $\mathbf{R^2}$, $i.e.$, $\{x\in\mathbf{R^2}|x^2_1+x^2_2=1\}$. Its affine hull is all of $\mathbf{R^2}$, so its affine dimension is two. By most definitions of dimension, however, the unit circle in $\mathbf{R^2}$ has dimension one.

I understood the affine hull of unit circle has dimension two because the all of values of affine hull is in $\mathbf{R^2}$. But still I don't understand why the unit circle in $\mathbf{R^2}$ has one dimension.

Thanks.

Michael Grant
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KimSJ
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    Do you believe that a line has dimension $1$ (by most reasonable definitions of dimension, at least)? If so, then a circle is locally the same as a line in a lot of meaningful senses. –  Jan 02 '14 at 04:36
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    $S^2$ can be represented by $e^{i\theta}$, which just needs one variable. – gaoxinge Jan 02 '14 at 05:31
  • @gaoxinge That would be $S^1$! – Bruno Joyal Feb 05 '14 at 04:10

2 Answers2

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You have to define dimension first, but intuitively the dimension is how many independent directions you can walk along if you were on the surface. On a circle, you can only walk back and forth along the circle, you don't have any other choice, so its dimension is $1$. More precisely, a circle is locally homeomorphic to a line, which has dimension $1$.

JLA
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I don't know what definition of dimension you are using, so I cannot give a precise answer. But let me say this: As a general principle, each new independent equation introduced in a system of equations cuts down the dimension of the solution space by $1$. This is true for systems of linear equations by elementary linear algebra, but it's also true for algebraic varieties in general, where the precise statement takes the form of the Krull principal ideal theorem.

Since $\mathbf R^2$ has dimension $2$, and the circle is determined by a single equation, it must have dimension $2-1=1$.

Bruno Joyal
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    This works well over algebraically closed field but not over the reals: A point in the plane is given by one algebraic equation $x^2+y^2=0$. You also need to invoke the fact that circle is given preimage of a regular value of one function. – Moishe Kohan Feb 14 '14 at 21:20
  • @studiosus Yes indeed. – Bruno Joyal Feb 14 '14 at 21:21